THE PREDUAL OF A VON NEUMANN ALGEBRA. A.P.M. KUPERS Contents 1. The predual 1 2. The case B(H): trace-class and Hilbert-Schmidt operators 2 3. Standard forms for II1 factors 5 We will always assume that our Hilbert spaces are separable. 1. The predual Let's fix some notation: if B is a Banach space, then B∗ is the vector space of norm-continuous ∗ linear functionals. This has two possible topologies: the operator norm topology (B)n and the ∗ weak-* topology (B)w. In general we will look at the latter. Let A be a von Neumann algebra. The goal of this talk will be naturally associate to a pair ∗ (A; A ,! B(H)) a Banach space A∗ such that A = (A∗)w, where the former is considered as having the ultraweak topology as a subspace of B(H). We claim that A∗ can be given by the space of ultraweakly continuous functionals. We will show ∗ ∼ that this is a Banach space and that (A∗)w = A. The Banach space A∗ will be called the predual. ∗ To get a better feeling for A∗, let A be the vector space of norm continuous linear functionals φ : A! C (with induced norm). Because norm convergence implies ultraweak convergence, it ∗ follows that A∗ is a subspace of A . We will reduce the proof to the case that A = B(H) using the following results. Recall that Hahn-Banach says that if one has a linear functional φ on a subspace of a vector space dominated by a sublinear functional N (think of it as a norm), then one can extend φ to the entire space in a way that it is still dominated by N . Proposition 1.1. If x 2 B(H) satisfies the φ(x) = 0 for each ultraweakly continuous φ with φ(A) = 0, then x 2 A. Proof. This works for V any ultraweakly closed subspace of B(H). If x2 = V , then apply Hahn- Banach or one of its corollaries (e.g. Conway IV.3.15) to the linear functional φ defined by φ(x) = 1 and φ(v) = 0 for all v 2 V . This is continuous with respect to the restriction of the ultraweak topology on V + Cx, hence extends to a ultraweakly continuous functional φ on B(H) with the property that φ(x) = 1. We prove the reduction to the case B(H) in the next section. However, we show that the embedding in B(H) is not essential for the theorem, by given an intrinsic definition of the predual. ∼ ∗ ∼ ? ∗ ? Theorem 1.2. Suppose that B(H)uw = (B)w. Then Auw = (B=A )w and furthermore B=A can be identified as a vector space with the subspace A∗ of ultraweakly continuous linear functionals on A. Date: March 20, 2011. 1 2 A.P.M. KUPERS ? ∗ Proof. We first identify Auw with (B=A )w. By definition there is an continuous embedding ∗ ∗ Auw ! B(H)uw = Bw. Because A is ultraweak closed, it is weak-* closed in B(H) = (B)w. This implies that there is a homeomorphism between the subspace of the norm-continuous functionals ? ∗ vanishing on A with the subspace topology of the weak-* one, and (B=A )w. This can be seen ? ∗ as follows: the embedding A! (B=A )w is clearly continuous, open and injective, and surjective using the previous proposition. Next we identify the vectorspace (B=A?)∗ with the space of ultraweakly continuous functionals ? on A. This means that we must identify the B=A with the weak-* continuous linear functionals A∗ ? ∗∗ ∗ ? in the double dual (B=A ) = (Auw) . See section V.1 of Conway: clearly the elements of B=A ∗ are weak-* continuous considered as elements of (Auw) . For the converse, one notes that a weak-* continous functional is bounded in absolute value by a finite sum of seminorms iv = b 7! jb(v)j for ? v 2 B=A and inductively use this to write the functional as a linear combination of the iv's. 2. The case B(H): trace-class and Hilbert-Schmidt operators In this section we will see that the predual of B(H) is given by the Banach space T (H) of traceclass operators with the trace-norm j − j1. Definition 2.1. Let a 2 B(H) and let (ξi) be an orthonormal basis. Then a is said to be of trace class if the following sum is finite X jaj1 = hjajξi; ξii i The set of trace-class operators is denoted by T (H). To show that this is well-defined, one proves that the sum is independent of the choice of orthonormal basis. Lemma 2.2. jaj1 is independent of the choice of orthonormal basis. Proof. Note that jaj = pjajpjaj and write X X p p hjajξi; ξii = h jajξi; jajξii i i X X p 2 = jh jajξi; ηiij i j X X p 2 = jh jajηi; ξiij j i X = hjajηj; ηji j where we were allowed to interchange the sum because all terms are positive. We now discuss several useful properties of the trace-class operators. Proposition 2.3. T (H) is an ideal which is closed under ∗. The map j − j1 is a complete norm on T (H) satisfying jjajj ≤ jaj1, making it into a Banach space. The finite rank operators are dense and hence T (H) is in fact a sub-ideal of the compact operators K(H). Proof. To show that it is an ideal in B(H), use that jxaj1 ≤ jjxjjjaj1 for x 2 B(H) and similarly ∗ jaxj1 ≤ jjxjjjaj1. That it is closed under ∗ is clear, as jaj = ja j. p 2 To show that jjajj ≤ jaj1 one remarks that jjajj ≤ jj jajjj ≤ jaj1. The first inequality follows p 2 from the functional calculus. For the second inequality, note that jj ajj = supjjξjj=1hjajξ; ξi and hence we pick any ξ of norm 1 for which this supremum is almost obtained as the first vector of p 2 an orthonormal basis to bound jaj1 ≥ jj jajjj − for all > 0. Taking to zero then proves the statement. To show that j − j1 is a complete norm, one notes that if an is j − j1-Cauchy, it is jj − jj-Cauchy to get a candidate a. We need to show that a is trace-class and ja−anj1 ! 0. However, the second statement implies the first since jaj1 ≤ janj1 + ja − anj1. So let's prove the second: if jja − anjj ! 0, THE PREDUAL OF A VON NEUMANN ALGEBRA. 3 then in particular jam − anjξi ! ja − anjξi for all vectors ξi of length 1. This implies that if P jam − anj1 = hjam − anjξi; ξii ≤ for m; n large enough, then ja − anj1 ≤ as well. Finally, note that any a 2 T (H) is the j − j1-limit of apn, where pn is the projection on the first n basis vectors ξi for 1 ≤ i ≤ n. All apn are of finite rank, hence these are dense. Because j − j1 dominates jj − jj, T (H) lies in the norm-closure of the finite-rank operators, which are the compact operators. Next we define the trace. Lemma 2.4. If a is a trace-class operator, then the following sum converges absolutely X Tr(a) = haξi; ξii i It satisfies Tr(a) ≤ jaj1 and Tr(xa) ≤ jjxjjTr(a) for each x 2 B(H). Proof. It suffices to prove the second statement. We will only prove the first part, as the second is then easy. To prove this, write a = upjajpjaj using the polar decomposition and consider the following: p p ∗ 2 0 ≤ jj( jaj − λ jaju )ξijj p 2 p p ∗ 2 p ∗ 2 = jj jajξijj − 2Re λh jajξi; jaju ξii + jλj jj jaju ξijj 2 ∗ ∗ = hjajξi; ξii − 2Re (λhaξi; ξii) + jλj hjaju ξi; u ξii iφ iφ Now write λ = e , where the phase is chosen such that Re e haξi; ξii = jhaξi; ξiij. Then we get: ∗ ∗ 2jhaξi; ξiij ≤ hjajξi; ξii + hjaju ξi; u ξii Now note that the sequences (ξi) and (uξi) both orthonormal bases. Thus summing over i we obtain Tr(a) ≤ jaj1 We will now prove that there is an isomorphism of Banach spaces between T (H)∗ with the operator norm topology and B(H) with the norm topology. An important role will be played by the map φ : B(H) !T (H)∗ given by φx(a) = Tr(xa) ∗ ∼ Theorem 2.5. We have that (T (H))n = B(H)n. Proof. By the previous proposition, we know that each φx is norm continuous. Thus, only the following statement is left: if λ : T (H) ! C is a linear functional which is bounded with respect to j − j1 then there is a x 2 B(H) such that λ(a) = φx(a) and jjλjj = jjxjj. Note that we have a sesquilinear form on B(H): hξ; ηiλ = λ(v 7! hv; ξiη) By Riesz representability, hξ; ηiλ = hxξ; ηi for some x 2 B(H) with jjxjj = sup jhξ; ηiλj = sup jλ(v 7! hv; ξiη)j = jjλjj jjξjj=jjηjj=1 jjξjj=jjηjj=1 where only the last equality is not obvious. Clearly the left hand side is less than or equal to the right hand side, because we should evaluate on more trace-class operators to get the usual definition. However, because jλ(−)j is continuous it suffices to check it on the dense subspace of finite rank operators. Linearity then implies we can restrict to trace-class operators of the form v 7! hv; ξiη if we want to show that the left hand side is greater than or equal to the right hand side.